Question: $\begin{cases}c(1)=\dfrac{3}{16}\\\\ c(n)=c(n-1)\cdot 4 \end{cases}$ What is the $3^{\text{rd}}$ term in the sequence?
Explanation: This is a recursive formula. It tells us that the first term is $\dfrac{3}{16}$ and that the common ratio is $4$. $\begin{aligned} {c(1)}&=\dfrac{3}{16} \\\\ {c(2)}&={c(1)}\cdot 4=\dfrac{3}{4} \\\\ {c(3)}&={c(2)}\cdot 4=3 \end{aligned}$ The $3^{\text{rd}}$ term is $3$.